This question is about the function \(Z\) defined by \( \displaystyle Z(s) = \sum_{k = 1}^\infty \frac{1}{k^s}, \) for \( s > 1 \).

1. Can you find a lower bound on this function which is valid for all \( s > 1 \)?
2. Why have we excluded \( s = 1\) from the domain?

From now on, we will concentrate on \( Z(2) \).

3. Can you find an upper bound on \(Z(2) \)?

In fact, \( \displaystyle Z(2) = \frac{\pi^2}{6} \), i.e., \( \displaystyle \sum_{k=1}^n \frac{1}{k^2} \rightarrow \frac{\pi^2}{6} \) as \( n \rightarrow \infty \).

4. One proof of this fact relies on the inequalities \( \sin \theta < \theta < \tan \theta,~~~~\)for \(\theta \in (0, \pi/2)\). Can you show why this is true geometrically?
5. Can you use this to get upper and lower bounds on \(1/\theta^2\) in terms only of \(\tan \theta\)?
6. Now let \( \theta_k = \frac{k \pi}{2 n + 1} \) for \(1 \leq k \leq n \). Find upper and lower bounds on \[ \sum_{k=1}^n \frac{1}{k^2}. \]
7. It turns out that the numbers \[ r_k = \frac{1}{\tan^2 \left( \frac{\pi k}{2n + 1 } \right) }, ~~ 1 \leq k \leq n, \] are roots of the polynomial \[ p_n(x) = { 2n + 1 \choose 1} x^n - {2n + 1 \choose 3} x^{n-1} + \cdots + (-1)^n {2n + 1 \choose 2n + 1}. \] Can you use this to show that \( \displaystyle \sum_{k=1}^n \frac{1}{k^2} \rightarrow \frac{\pi^2}{6} \) as \( n \rightarrow \infty \)?

This question was used in Maths & CS interviews at Lady Margaret Hall in December 2020. The question is due to Christina Goldschmidt.